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## A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge at point P, w

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## Answers ( )

Answer:E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

(1)Where the minus sign means that the electric field point to the charge.

k: Coulomb’s constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

The distance r is:

You replace the values of all parameters in the equation (1):

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C